Question

具有以下import mysql.connector dbname = input('Please enter the name of database : ') db = mysql.connector.connect( host = 'localhost', user = 'root', password = '', ) cursor = db.cursor() cursor.execute("CREATE DATABASE IF NOT EXISTS %s" %dbname) cursor.close() db.close() import mysql.connector db = mysql.connector.connect( host = 'localhost', user = 'root', password = '', database=dbname, ) cursor = db.cursor() cursor.execute("CREATE TABLE IF NOT EXISTS %s (email VARCHAR(30),pwd VARCHAR(20))" %dbname) cursor.close() db.close()

pd.DataFrame

我想对行中的连续零进行计数

pd.DataFrame({'2010':[0, 45, 5], '2011': [12, 56, 0], '2012': [11, 22, 0], '2013': [0, 5, 0], '2014': [0, 0, 0]})

  2010 2011 2012 2013 2014
1  0    12   11   0    0
2  45   56   22   5    0
3  5    0    0    0    0

寻找不同的有效方式

Answer 1

为了提高效率，我建议您采用纯NumPy方式-

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  <assemblyIdentity name="System.Web.Http" publicKeyToken="31bf3856ad364e35" xmlns="urn:schemas-microsoft-com:asm.v1" />
  <bindingRedirect oldVersion="0.0.0.0-5.2.3.0" newVersion="5.2.3.0" xmlns="urn:schemas-microsoft-com:asm.v1" />
</dependentAssembly>

样品运行-

def islandlen_perrow(df, trigger_val=0):
    a=df.values==trigger_val
    pad = np.zeros((a.shape[0],1),dtype=bool)
    mask = np.hstack((pad, a, pad))
    mask_step = mask[:,1:] != mask[:,:-1]
    idx = np.flatnonzero(mask_step)
    island_lens = idx[1::2] - idx[::2]
    n_islands_perrow = mask_step.sum(1)//2
    out = np.split(island_lens,n_islands_perrow[:-1].cumsum())
    return out

大型数组上的计时-

In [69]: df
Out[69]: 
   2010  2011  2012  2013  2014
0     0    12    11     0     0
1    45    56    22     5     0
2     5     0     0     0     0

In [70]: islandlen_perrow(df, trigger_val=0)
Out[70]: [array([1, 2], dtype=int64), array([1], dtype=int64), array([4], dtype=int64)]

In [76]: pd.Series(islandlen_perrow(df, trigger_val=0))
Out[76]: 
0    [1, 2]
1       [1]
2       [4]
dtype: object

Answer 2

您可以使用itertools.groupby：

import pandas as pd

from itertools import groupby


def count_zeros(x):
    return [sum(1 for _ in group) for key, group in groupby(x, key=lambda i: i == 0) if key]


df = pd.DataFrame({'2010':[0, 45, 5], '2011': [12, 56, 0], '2012': [11, 22, 0], '2013': [0, 5, 0], '2014': [0, 0, 0]})

result = df.apply(count_zeros, axis=1)
print(result)

输出

0    [1, 2]
1       [1]
2       [4]
dtype: object

Answer 3

将itertools.groupby与列表理解结合使用：

from itertools import groupby

df['counts'] = [[len(list(grp)) for flag, grp in groupby(row, key=bool) if not flag] \
                for row in df.values]

print(df)

   2010  2011  2012  2013  2014  counts
0     0    12    11     0     0  [1, 2]
1    45    56    22     5     0     [1]
2     5     0     0     0     0     [4]

Answer 4

如果您对纯熊猫/ numpy解决方案感兴趣...可以使用groupby和value_counts：

v = df.stack()
m = v.eq(0)

(m.ne(m.shift())
  .cumsum()
  .where(m)
  .dropna()
  .groupby(level=0)
  .apply(lambda x: x.value_counts(sort=False).tolist()))

0    [1, 2]
1       [1]
2       [4]
dtype: object

或者，避免使用lambda，

(m.ne(m.shift())
  .cumsum()
  .where(m)
  .dropna()
  .groupby(level=0)
  .value_counts(sort=False)
  .groupby(level=0)
  .apply(list))

0    [1, 2]
1       [1]
2       [4]
dtype: object

Answer 5

一种方法是将值转换为布尔值，然后用1 [1, 2] 2 [1] 3 [4]值分割字符串

False

计算熊猫行上的连续零

5 个答案: